Answer:
Step-by-step explanation:
Hello!
You have the data for 4 observations of randomly selected employees of the U.S.
Wage: 16.20; 12.36; 14.40; 12.00
Education: 12; 13; 12; 12
X₁: Wage of a U.S. employee.
X₂: Education of a U.S. employee.
i.
You have to estimate with a 95% confidence the population mean of each variable. Assuming all conditions are met, I'll use a students t to estimate both means.
* The general formula for the CI is: [X[bar]±
*
]
![t_{n-1;1-\alpha /2}= t_{3;0.975}= 3.182](https://tex.z-dn.net/?f=t_%7Bn-1%3B1-%5Calpha%20%2F2%7D%3D%20t_%7B3%3B0.975%7D%3D%203.182)
<u>For population 1</u> (Wages)
n₁= 4; X[bar]₁= 13.74; S₁= 1.95
13.74±3.182*
]
[10.65; 16.85]
Using a 95% confidence level you'd expect that the interval [10.65; 16.85] contains the population mean of the wages of U.S. employees.
<u>For population 2 </u>(Education)
n₂= 4; X[bar]₂= 12.25; S₂= 0.50
12.25±3.182*
]
[14.45; 13.05]
Using a 95% confidence level you'd expect that the interval [14.45; 13.05] contains the value of the average education of U.S. employees.
ii.
To estimate Rho (the population correlation coefficient) you have to use the following formula:
![r= \frac{sumX_1X_2-\frac{(sumX_1)(sumX_2)}{n} }{\sqrt{[sumX_1^2-\frac{(sumX_1)^2}{n_1} ][sumX_2^2-\frac{(sumX_2)^2}{n_2} ]} }](https://tex.z-dn.net/?f=r%3D%20%5Cfrac%7BsumX_1X_2-%5Cfrac%7B%28sumX_1%29%28sumX_2%29%7D%7Bn%7D%20%7D%7B%5Csqrt%7B%5BsumX_1%5E2-%5Cfrac%7B%28sumX_1%29%5E2%7D%7Bn_1%7D%20%5D%5BsumX_2%5E2-%5Cfrac%7B%28sumX_2%29%5E2%7D%7Bn_2%7D%20%5D%7D%20%7D)
∑X₁= 54.96; ∑X₁²= 766.57; ∑X₂= 49; ∑X₂²= 601; ∑X₁X₂= 671.88
![r= \frac{671.88-\frac{54.96*49}{4} }{\sqrt{[766.57-\frac{(54.96)^2}{4} ][601-\frac{(49)^2}{4} ]} }](https://tex.z-dn.net/?f=r%3D%20%5Cfrac%7B671.88-%5Cfrac%7B54.96%2A49%7D%7B4%7D%20%7D%7B%5Csqrt%7B%5B766.57-%5Cfrac%7B%2854.96%29%5E2%7D%7B4%7D%20%5D%5B601-%5Cfrac%7B%2849%29%5E2%7D%7B4%7D%20%5D%7D%20%7D)
r= -0.47
iii.
To estimate the regression using Y: wage and X: education you have to estimate the intercept and the slope of the equation:
<u>Estimate of the slope "b"</u>
![b= \frac{sumXY-\frac{(sumX)(sumY)}{n} }{[sumX^2-\frac{(sumX)^2}{n} ]} = \frac{671.88-\frac{54.96*49}{4} }{601-\frac{49^2}{4} } = -1.84](https://tex.z-dn.net/?f=b%3D%20%5Cfrac%7BsumXY-%5Cfrac%7B%28sumX%29%28sumY%29%7D%7Bn%7D%20%7D%7B%5BsumX%5E2-%5Cfrac%7B%28sumX%29%5E2%7D%7Bn%7D%20%5D%7D%20%3D%20%5Cfrac%7B671.88-%5Cfrac%7B54.96%2A49%7D%7B4%7D%20%7D%7B601-%5Cfrac%7B49%5E2%7D%7B4%7D%20%7D%20%3D%20-1.84)
<u>Estimate of the intercept "a"</u>
![a= Y[bar] - b*X[bar]= 13.74 - (-1.84)*12.25= 36.28](https://tex.z-dn.net/?f=a%3D%20Y%5Bbar%5D%20-%20b%2AX%5Bbar%5D%3D%2013.74%20-%20%28-1.84%29%2A12.25%3D%2036.28)
The estimated regression equation is ^Y= 36.28 - 1.84X
iv.
You have to estimate the value of the wages ^Y given X= 15years of education. To do so you have to replace the value of X in the estimated regression equation:
^Y= 36.28 - 1.84*15= 8.68
For a level of education of 15 years, the estimated wage is 8.68
v. The value of the coefficient of determination is R²= 0.22
This means that 22% of the variability of the average wages of U.S. employees is explained by the years of education. For the estimated model ^Y= 36.28 - 1.84X.
I hope this helps!